In this instance, we will explore a C++ code to identify a matrix containing highlighted regions marked by asterisks.
Problem Statement:
Assume there is an n × n grid consisting of asterisks (*) and dots (.). Almost every cell contains a dot except for two cells. These two cells are required to be identified as the corners to form a rectangle with sides that are parallel to the coordinate axes.
Overview:
The algorithm iterates over the matrix to find two asterisks that constitute a region. Next, it expands the area by annotating neighboring cells according to how close they are to the original asterisks. The extension follows a simple set of guidelines:
- Cells below or above the asterisks are indicated if they are in the same row.
- Cells to their left or right are marked if they are in the same column.
- Cells diagonally opposite the asterisks are marked if they are not.
Algorithm:
function mark(mat)
total:= size of mat
a := -1, b := -1, c := -1, d := -1
for i:= 0 to total - 1,then do
for j:= 0 to total - 1, then do
if mat[i][j] == '*' and a == -1 then
a := i
c := j
else if mat[i][j] == '*' and a != -1 then
b := i
d := j
if a == b then
if a != total - 1 then
mat[a + 1][c] := '*', mat[b + 1][d] := '*'
else
mat[a - 1][c] := '*', mat[b - 1][d] := '*'
else if c == d then
if c != total - 1 then
mat[a][c + 1] := '*', mat[b][d + 1] := '*'
else
mat[a][c - 1] := '*', mat[b][d - 1] := '*'
else
mat[a][d] := '*', mat[b][c] := '*'
function main()
total:= input integer from the user
mat := create 2D array of characters with size total * total
for i:= 0 to total - 1, then do
for j:= 0 to total - 1, then do
mat[i][j] := input character from user
call mark(mat)
for i:= 0 to total - 1, then do
for j:= 0 to total - 1, then do
print mat[i][j] followed by a space
Example:
Let's consider a C++ program that identifies a matrix containing a highlighted region of asterisks.
#include<iostream>
#include<vector>
using namespace std;
void mark(vector<vector<char>>& mat)
{
int num = mat.size();
int , b, c, d, m, n;
int a = b = c = d = -1;
// Find the coordinates of the asterisks in the matrix
for (int m = 0; n < num; ++m) {
for (int m = 0; n < num; ++n) {
if (mat [m][n] == '*' && a == -1) {
a = m;
c = n;
}
else if (mat [m][n] == '*' && a != -1) {
b = m;
d = n;
}
}
}
// Mark adjacent cells
if (a == b) {
if (a != num - 1) {
mat [a + 1][c] = mat [b + 1][d] = '*';
}
else {
mat [a - 1][c] = mat [b - 1][d] = '*';
}
}
else if (c == d) {
if (c != num - 1) {
mat [a][c + 1] = mat [b][d + 1] = '*';
}
else {
mat [a][c - 1] = mat [b][d - 1] = '*';
}
}
else {
mat [a][d] = mat [b][c] = '*';
}
}
int main() {
int num;
cout << "Enter the size of the matrix: ";
cin >> num;
vector<vector<char>> mat (num, vector<char>(num));
cout << "Enter the elements of the matrix "<< endl;
for (int m = 0; m < num; ++m) {
for (int n = 0; n < num; ++n) {
cin >> mat [m][n];
}
}
mark(mat);
cout << "Matrix with marked asterisks region:" << endl;
for (int m = 0; m < num; ++m) {
for (int n = 0; n < num; ++n) {
cout << mat [m][n] << " ";
}
cout << endl;
}
return 0;
}
Output:
Complexity Analysis:
Time Complexity:
To locate the asterisks within the matrix, an algorithm with a time complexity of O(n^2) is used.
Space Complexity:
O(n^2) - It depends on the size of the matrix
Conclusion:
In summary, the provided C++ code efficiently identifies a matrix area containing asterisks by systematically discovering and enlarging the region based on specific criteria. This straightforward method serves as a valuable asset for analyzing matrices and recognizing patterns, as it can be implemented across diverse computational scenarios.
